91Ó°ÊÓ

The given data can be listed below as,

• The coordinate of the location A is, $\mathrm{A}⟨{\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1âŸ\copyright =⟨-0.5,0,0âŸ\copyright \text{ }\mathrm{m}$.
• The coordinate of the location B is,$\mathrm{B}⟨{\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2âŸ\copyright =⟨0.5,0,0âŸ\copyright \text{ }\mathrm{m}$.
• The electric in a region is,$⟨{\mathrm{E}}_{\mathrm{x}},{\mathrm{E}}_{\mathrm{y}},{\mathrm{E}}_{\mathrm{z}}âŸ\copyright =⟨750,0,0âŸ\copyright \text{ }\mathrm{N}/\mathrm{C}$.

In this question, the potential energy change for a charged particle system can be obtained with the help of the particle's electric charge and potential difference. The potential energy change depends directly on the electric charge of the particle.

The relation ofdisplacement vector$\mathrm{Δ}\stackrel{→}{\mathrm{l}}$is expressed as,

$\mathrm{Δ}\stackrel{→}{\mathrm{l}}=⟨({\mathrm{x}}_2−{\mathrm{x}}_1),({\mathrm{y}}_2−{\mathrm{y}}_1),({\mathrm{z}}_2-{\mathrm{z}}_1)âŸ\copyright$

Here, $\mathrm{Δ}\stackrel{→}{\mathrm{l}}$is the displacement vector.

Substitute all the known values in the above equation.

$\begin{array}{c}\mathrm{Δ}\stackrel{→}{\mathrm{l}}=⟨\{0.5\text{″¾}−(−0.5\text{″¾})\},(0−0),(0-0)âŸ\copyright \\ =⟨1,0,0âŸ\copyright \text{″¾}\end{array}$

Thus, the displacement vector $\mathrm{Δ}\stackrel{→}{l}$ is $⟨1,0,0âŸ\copyright \text{″¾}$.

The relation ofchange in electric potentialis expressed as,

$\mathrm{Δ}\stackrel{→}{\mathrm{V}}=−⟨{\mathrm{E}}_{\mathrm{x}}â‹\dots ({\mathrm{x}}_2−{\mathrm{x}}_1),{\mathrm{E}}_{\mathrm{y}}â‹\dots ({\mathrm{y}}_2−{\mathrm{y}}_1),{\mathrm{E}}_{\mathrm{z}}â‹\dots ({\mathrm{z}}_2−{\mathrm{z}}_1)âŸ\copyright$

Here, $\mathrm{Δ}\stackrel{→}{\mathrm{V}}$is the change in electric potential.

Substitute all the known values in the above equation.

$\begin{array}{c}\mathrm{Δ}\stackrel{→}{V}=−⟨\{\left(750\text{ N/°ä}\right)â‹\dots (0.5\text{″¾â€‰}+0.5\text{″¾})\},\{\left(0\text{ N/°ä}\right)â‹\dots (0−0)\},\{\left(0\right)â‹\dots (0−0)\}âŸ\copyright \\ =−⟨750\text{ N}â‹\dots \text{³¾/°ä, 0 ,0}âŸ\copyright \left(\frac{1\text{ V}}{1\text{ N}â‹\dots \text{m/C}}\right)\\ =−⟨750\text{, 0 ,0}âŸ\copyright \text{ V}\end{array}$

Thus, the change in electric potential is $−⟨750\text{, 0 ,0}âŸ\copyright \text{ V}$.

The relation of potential energy change for the system when a proton moves from B to A is expressed as,

$\begin{array}{c}{\left(\mathrm{Δ\pm «}\right)}_{\mathrm{p}}={\mathrm{q}}_{\mathrm{p}}â‹\dots \left(\mathrm{Δ}\stackrel{→}{\mathrm{V}}\right)\\ =−{\mathrm{q}}_{\mathrm{p}}⟨{\mathrm{E}}_{\mathrm{x}}â‹\dots ({\mathrm{x}}_1−{\mathrm{x}}_2),{\mathrm{E}}_{\mathrm{y}}â‹\dots ({\mathrm{y}}_1−{\mathrm{y}}_2),{\mathrm{E}}_{\mathrm{z}}â‹\dots ({\mathrm{z}}_1−{\mathrm{z}}_2)âŸ\copyright \end{array}$

Here, ${\left(\mathrm{Δ\pm «}\right)}_{\mathrm{p}}$is the potential energy change for the system when a proton moves from B to Aand ${\mathrm{q}}_{\mathrm{p}}$ represents the charge on a proton whose value is $1.6×{10}^{−19}\text{ C}$.

Substitute all the known values in the above equation.

$\begin{array}{c}{\left(\mathrm{Δ\pm «}\right)}_{\mathrm{p}}=−(1.6×{10}^{−19}\text{ C})â‹\dots ⟨\{\left(750\text{ N/°ä}\right)â‹\dots (−0.5\text{″¾}−0.5\text{″¾})\},\{\left(0\right)â‹\dots (0−0)\},\{\left(0\right)â‹\dots (0−0)\}âŸ\copyright \\ =−(1.6×{10}^{−19}\text{ C})â‹\dots ⟨−750,0,0\text{ }âŸ\copyright \text{ N}â‹\dots \text{m/C}\\ =⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ N}â‹\dots \text{³¾â€‰}×\left(\frac{1\text{ J}}{1\text{ N}â‹\dots \text{m}}\right)\\ =⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ J}\end{array}$

Thus, the potential energy change for the system when a proton moves from B to A is$⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ J}$ .

The relation of potential energy change for the system when an electron moves from B to A is expressed as,

$\begin{array}{c}{\left(\mathrm{Δ\pm «}\right)}_{\mathrm{e}}={\mathrm{q}}_{\mathrm{e}}â‹\dots \left(\mathrm{Δ}\stackrel{→}{\mathrm{V}}\right)\\ =−{\mathrm{q}}_{\mathrm{e}}⟨{\mathrm{E}}_{\mathrm{x}}â‹\dots ({\mathrm{x}}_1−{\mathrm{x}}_2),{\mathrm{E}}_{\mathrm{y}}â‹\dots ({\mathrm{y}}_1−{\mathrm{y}}_2),{\mathrm{E}}_{\mathrm{z}}â‹\dots ({\mathrm{z}}_1−{\mathrm{z}}_2)âŸ\copyright \end{array}$

Here,${\left(\mathrm{Δ\pm «}\right)}_{\mathrm{e}}$ is the potential energy change for the system when an electron moves from B to Aand ${\mathrm{q}}_{\mathrm{p}}$ represents the charge on an electron whose value is $−1.6×{10}^{−19}\text{ C}$.

Substitute all the known values in the above equation.

$\begin{array}{c}{\left(\mathrm{Δ}U\right)}_e=−(−1.6×{10}^{−19}\text{ C})â‹\dots ⟨\{\left(750\text{ N/°ä}\right)â‹\dots (−0.5\text{″¾}−0.5\text{″¾})\},\{\left(0\right)â‹\dots (0−0)\},\{\left(0\right)â‹\dots (0−0)\}âŸ\copyright \\ =−(−1.6×{10}^{−19}\text{ C})â‹\dots ⟨−750,0,0\text{ }âŸ\copyright \text{ N}â‹\dots \text{m/C}\\ =−⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ N}â‹\dots \text{³¾â€‰}×\left(\frac{1\text{ J}}{1\text{ N}â‹\dots \text{m}}\right)\\ =−⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ J}\end{array}$

Thus, the potential energy change for the system when an electron moves from B to A is $−⟨1.2×{10}^{−16},0,0âŸ\copyright \text{ J}$.